#### provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18.9 question 62

Answer: $\frac{\sec ^{3} 2 x}{6}-\frac{\sec 2 x}{2}+c$

Hint: Use substitution method to solve this integral

Given: $\int \tan ^{3} 2 \: x \cdot \sec 2 \: x \; d x$

Solution:

$\text { Let } I=\int \tan ^{3} 2 x \cdot \sec 2 x\; d x$

$\Rightarrow I=\int \tan ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x\; d x$

\begin{aligned} &=\int\left(\sec ^{2} 2 x-1\right) \tan 2 x \cdot \sec 2 x \; d x \\ &=\int\left(\sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x-\tan 2 x \sec 2 x\right) d x \\ &=\int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x\; d x-\int \tan 2 x \cdot \sec 2 x\; d x \end{aligned}

\begin{aligned} &\Rightarrow I=\int \sec ^{2} 2 x \cdot \tan 2 x \cdot \sec 2 x \; d x-\int \tan 2 x \cdot \sec 2 x\; d x \\ &\text { Put } \sec 2 x=t \Rightarrow \sec 2 x . \tan 2 x \cdot 2 \cdot d x=d t \end{aligned}

$\Rightarrow d x=\frac{d t}{2 \sec 2 x \cdot \tan 2 x} \text { then the Ist integral becomes }$

$\Rightarrow I=\int t^{2} \cdot \tan 2 x \cdot \sec 2 x \cdot \frac{d t}{2 \sec 2 x \cdot \tan 2 x}-\int \sec 2 x \cdot \tan 2 x \; d x$

\begin{aligned} &=\int \frac{t^{2}}{2} d t-\int \sec 2 x \cdot \tan 2 x\; d x \\ &=\frac{1}{2} \int t^{2} d t-\int \sec 2 x \cdot \tan 2 x\; d x=\frac{1}{2} \frac{t^{2+1}}{2+1}-\frac{\sec 2 x}{2}+c \end{aligned}

$=\frac{1}{2} \cdot \frac{t^{3}}{3}-\frac{\sec 2 x}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c, \int \sec a x \cdot \tan a x \; d x=\frac{\sec a x}{a}+c\right]$

$=\frac{1}{6} \sec ^{3} 2 x-\frac{\sec 2 x}{2}+c \quad[\because t=\sec 2 x]$